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  • A particle of mass m is droped inside a liquid medium which apply a bouyancy force V and a drag force <img alt="\mathrm{kv^{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

A particle of mass m is droped inside a liquid medium which apply a bouyancy force V and a drag force \mathrm{kv^{2}}, if acceleration due to gravity \mathrm{g} is constant. what is terminal velocity attaint by the particle -

Option: 1

\mathrm{\sqrt{\frac{m g+B}{k}}}


Option: 2

\mathrm{\sqrt{\frac{m g+2 B}{k}}}


Option: 3

\mathrm{\sqrt{\frac{m g-B}{k}}}


Option: 4

\mathrm{\frac{m g-B}{k}}


Answers (1)

best_answer

At \mathrm{ v=v_t--- terminal \: \: velocity}

From figure, 

\mathrm{\begin{aligned} & m g=B+K V_t^2 \\ & m g-B=k V_t^2 \\ & V_t^2=\frac{m g-B}{k} \\ & V_t=\sqrt{\frac{m g-B}{k}} \end{aligned}}

\therefore Terminal velocity is always opposite to velocity.

 

Posted by

Ritika Jonwal

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